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VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
15 tháng 6 2022
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
11 tháng 7 2018
\(x^3+x=0\)
\(\Rightarrow x.\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\end{cases}}\)
\(x^2-2x-3=0\)
\(\Rightarrow x.\left(x-2\right)=3\)
Vì \(x>x-2\)và \(x\inƯ\left(3\right)=\left\{3;-3\right\}\)
Các phần sau tương tự
KT
11 tháng 7 2018
\(x^3+x=0\)
\(\Leftrightarrow\)\(x\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
\(x^2-2x-3=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
\(x+5x^2=0\)
\(\Leftrightarrow\)\(x\left(5x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
Vậy...
MN
4
11 tháng 7 2018
\(b,x^2-6x+8=0\)
\(\Rightarrow x^2-2.3.x+9=1\)
\(\Rightarrow\left(x-3\right)^2=1\)
\(\Rightarrow x-3=\orbr{\begin{cases}1\\-1\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}4\\2\end{cases}}\)
\(d,x+5x^2=0\)
\(\Rightarrow x\left(1+5x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
KT
11 tháng 7 2018
a) \(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
b) \(x^2-6x+8=0\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
Vậy...
c) \(x+5x^2=0\)
\(\Leftrightarrow\)\(x\left(5x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
Vậy...
16 tháng 5 2018
\(\dfrac{2}{2x-6}+\dfrac{2}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\) ( x # 3 ; x # -1)
⇔ \(\dfrac{2}{2\left(x-3\right)}+\dfrac{2}{2\left(x+1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
⇔ \(\dfrac{x+1}{\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{\left(x+1\right)\left(x-3\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
⇔ x + 1 + x - 3 - 2x = 0
⇔ - 2 = 0 ( vô lý )
Vậy , phương trình vô nghiệm
25 tháng 6 2019
Tìm x,biết:
a/ x + 5x2 =0
⇔x ( 1 + 5x ) = 0
\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0
1) x = 0
2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)
b/x+1=(x+1)2
\(\Leftrightarrow\) (x+1) - (x+1)2 = 0
\(\Leftrightarrow\) ( x+ 1)(1-x-1) = 0
\(\Leftrightarrow\) (x+1).(-x) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x = 0
\(\Leftrightarrow\) x= -1 ; 0
Vậy: S=\(\left\{-1;0\right\}\)
c/ x3+x=0
\(\Leftrightarrow\) x(x2 + 1) = 0
\(\Leftrightarrow\) x = 0 hoặc x2 + 1 = 0
Ta có : x2 + 1 \(\ge\) 0 vs mọi x
Vậy: S = \(\left\{0\right\}\)
d/5x(x−2)−(2−x)=0
\(\Leftrightarrow\) 5x(x-2) + (x - 2) = 0
\(\Leftrightarrow\) (x - 2)(5x+1) = 0
\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0
\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)
Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)
g/ x(x−4)+(x−4)2=0
⇔ (x - 4)( x+x-4) = 0
\(\Leftrightarrow\) x - 4 = 0 hoặc 2x-4=0
\(\Leftrightarrow\) x = 4 hoặc x = 2
Vậy: S= \(\left\{2;4\right\}\)
h/ x2−3x=0
⇔x (x-3) = 0
\(\Leftrightarrow\) x = 0 hoặc x = 3
Vậy: S = \(\left\{0;3\right\}\)
Vậy: S= \(\left\{0;3\right\}\)
i/4x(x+1)=8(x+1)
⇔4x(x+1)-8(x+1) = 0
\(\Leftrightarrow\) 4(x+1) (x - 2) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x= -1 hoặc x = 2
Vậy: S=\(\left\{-1;2\right\}\)
28 tháng 12 2017
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
28 tháng 12 2017
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
x2+5x-8=0
x(x+5)-8=0
x(x+5)=0+8
x(x+5)=-8
* x=-8
*x+5=-8 => x=-13
vậy...........................
toán 9 à bạn ? ^^
\(x^2+5x-8=0\)
Ta có : \(\Delta=5^2-4\left(-8\right)=25+32=57\)
do \(\Delta>0\)nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{-5+\sqrt{57}}{2}\)\(x_2=\frac{-5-\sqrt{57}}{2}\)
vậy ...